What is the general solution for 3=4sin(8(-pi/2-x))+3 ?

The general solution for 3=4sin(8(-pi/2-x))+3 is x=-(pin)/4-pi/2 ,x=-(5pi+2pin)/8

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3=4sin(8(-pi/2-x))+3

3 = 4 sin (8 (- \frac{π}{2} - x)) + 3

x = - \frac{πn}{4} - \frac{π}{2}, x = - \frac{5 π + 2 πn}{8}

Solution steps

3 = 4 sin (8 (- \frac{π}{2} - x)) + 3

Switch sides

4 sin (8 (- \frac{π}{2} - x)) + 3 = 3

Subtract

3

from both sides

4 sin (8 (- \frac{π}{2} - x)) + 3 - 3 = 3 - 3

Simplify

4 sin (8 (- \frac{π}{2} - x)) = 0

Divide both sides by

4

sin (8 (- \frac{π}{2} - x)) = 0

General solutions for

sin (8 (- \frac{π}{2} - x)) = 0

8 (- \frac{π}{2} - x) = 0 + 2 πn, 8 (- \frac{π}{2} - x) = π + 2 πn

Solve

8 (- \frac{π}{2} - x) = 0 + 2 πn : x = - \frac{πn}{4} - \frac{π}{2}

Solve

8 (- \frac{π}{2} - x) = π + 2 πn : x = - \frac{5 π + 2 πn}{8}

x = - \frac{πn}{4} - \frac{π}{2}, x = - \frac{5 π + 2 πn}{8}

5 sin (x) - 7 = 0

tan (\frac{θ}{2} + \frac{π}{4}) = - 1, 0 \leq θ \leq 2 π

sin^{2} (x) = 2 cos (x) + 2, 0 \leq x \leq 2 π

tan^{4} (x) = tan^{4} (x) cos (x)

tan (x) sec (x) = 0

What is the general solution for 3=4sin(8(-pi/2-x))+3 ?
The general solution for 3=4sin(8(-pi/2-x))+3 is x=-(pin)/4-pi/2 ,x=-(5pi+2pin)/8